In dealing with continued fractions in LaTeX, I've become less and less enamored of the \cfrac construct. Especially in my case where I deal with complicated numerators and denominators, listing a few terms can get unwieldy.

I encountered a nice notation for CFs in Lorentzen and Waadeland's "Continued Fractions with Applications" that seemed to be adaptable to my needs. I have no digital copy of the book, so I offer Wolfram's approximation on how this operator is rendered:

(the only difference is that the b terms come before the a terms, and they are separated by a backslash (this is why I asked this question, so I now know how to render the operands and the delimiters properly))

My question now, then, is how can I construct a "continued fraction operator" that acts like \sum and \product? Somehow, trying to use a large letter K and then putting the limits as over- and underscripts has been a bit of a mixed bag (and of course, if the operands themselves are fractions, I have to manually tweak the size of the "K" again!). Might there be a more elegant approach to this?

Wikipedia says the notation is due to Gauss. Despite its distinguished origin, I don't like it: I have to consciously resist cancelling the fraction. Better to add some sign to the fraction, say a trailing vertical slash on the denominator, to indicate this isn't a real fraction.
–
Charles StewartAug 24 '10 at 10:01

That's actually why I substitute a backslash for a slash in my set of notes, and then place b before a; at least for me it reinforces psychologically that this is not your "typical" fraction. I'm of course not entirely sure that my modification will catch on, but what can you do... :)
–
user914Aug 24 '10 at 12:36

The documentation for amsmath says \operatorname* gives this behavior but doesn't mention \operatornamewithlimits at all. Is this an undocumented synonym?
–
Mark MeckesAug 24 '10 at 17:26

You know, I'm not sure where I picked up this trick. I use it for argmax and argmin.
–
GeoffAug 24 '10 at 20:45

Egh, I guess I was unclear! I said the notation at Wolfram was just an approximation! (What I was trying to do looks something like K(b\a) with the appropriate over- and underscripts.) I think I can use this, however. Thanks!
–
user914Aug 24 '10 at 22:07

To use your \bigk, the notation I would actually be using would be something like \[ \bigk_{k=0}^\infty \bigl (b_k \bigm\backslash a_k \bigr ) \] ; would this scale properly if, say, I replace the a_k and b_k with a \frac construct?
–
user914Aug 23 '10 at 6:17

@J. Mangaldan: Do \sum and \product scale automatically? I don't see that they do?
–
Lev BishopAug 23 '10 at 6:30

2

I would use \vcenter{\hbox{...}} rather than \raisebox{-5pt}{...}
–
Lev BishopAug 23 '10 at 6:31

Lev: Odd... I tried it out: latex.codecogs.com/gif.latex?\sum_{j=0}^{\infty}\frac{\frac{\ln(j+1)}{j!}‌​}{\binom{n}{j}} just now... I suppose my memory was mistaken. So using that construct would make the K act like \sum, \product, and \int?
–
user914Aug 23 '10 at 6:39